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You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.
Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.
Example 1:
Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.Example 2:
Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
Total amount you can rob = 2 + 9 + 1 = 12.
方法1：
public class Leetcode_198_HouseRobber_DP1 {
public static void main(String[] args) {
Leetcode_198_HouseRobber_DP1 object = new Leetcode_198_HouseRobber_DP1();
int[] nums = new int[]{2, 7, 9, 3, 1};
// int[] nums = new int[]{};
System.out.println(object.rob(nums));
}
public int rob(int[] nums) {
if (nums.length == 0) {//[]不等于null，需要用长度判断
return 0;
}
if (nums != null && nums.length == 1) {
return nums[0];
}
if (nums.length == 2) {
Math.max(nums[0], nums[1]);
}
int[] dp = new int[nums.length];
dp[0] = nums[0];
dp[1] = Math.max(nums[0], nums[1]);//因为nums[0]与nums[1]均可以与nums[3]组合，所以取两者的最大值
for (int i = 2; i < nums.length; i++) {
dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);
}
return dp[nums.length - 1];
}
}输出12
Runtime: 4 ms, faster than 41.21% of Java online submissions for House Robber.
end